Question

convergence of a series.

Answer 1

\[\Large \sum_{n=1}^{\infty}\left(\frac{3 \cdot 2^n}{6^{n-1}} \right)\]\[\Large \sum_{n=1}^{\infty}\left(\frac{3 \cdot 2^n}{6^{n}\cdot6^{-1}} \right)\]\[\Large \left(\frac{3}{6^{-1}} \right) \times \sum_{n=1}^{\infty}\left(\frac{2^n}{6^{n}} \right)\]\[\Large \left(18\right) \times \sum_{n=1}^{\infty}\left(\frac{1}{3} \right)^n \]

Answer 2

So my sum would be\[\large 18 \times \left[ \frac{a_1}{1-r}\right]\]\[\large 18 \times \left[ \frac{6}{1-(1/3)}\right]\]\[18 \times 6 \div 2 \times 3= ...\]

Answer 3

@jdoe0001 @iambatman so the series is going to be converging to whatever that is?

Answer 4

(which is 162 )

Answer 5

@FibonacciChick666 can you check my work ?

Answer 6

So, just a thing, state whichever test you used

Answer 7

for convergence of a sequence (not a sum - series) I simplified the series and it comes out to be that ratio is 1/3 - that is the absolute value of ratio is less than 1.

Answer 8

I apologize for not using calculus on this one. But did I really have to do so in this case?

Answer 9

just for convergence/divergence I would have used the nth root test

Answer 10

but, let me check it the way you did, your simplification is correct

Answer 11

isn't the limit of 1/3^n zero? so shouldn't that sum be like 1?

Answer 12

the terms approach zero, but that doesn't say anything about the sum.
or am I wrong ?

Answer 13

I use it to guage what the sum will equal so currently it's 1/3+1/9+1/27 ...

Answer 14

well, my first term is 6, and ratio of 1/3

Answer 15

Is that ever going to add to 1 is my question

Answer 16

your first term is 1/3

Answer 17

\[\frac{3 \times 2^\color{red}{1}}{6^{1-\color{red}{1}}}=\frac{6}{6^0}=6/1=6\]@FibonacciChick666

Answer 18

... no, you have \[\Large \left(18\right) \times \sum_{n=1}^{\infty}\left(\frac{1}{3} \right)^n\]

Answer 19

oh, without multiplying times, 18 taking the simpler version of (1/3)^n ...

Answer 20

use your simplified series, yea

Answer 21

I see what you are doing.... I was going from the beginn

Answer 22

yea, I know

Answer 23

it's ok, so anyways, you find the value of that series then multiply by 18 and voila you have beaten the problem :)

Answer 24

ok,
1/3 +1/9 = 3/9 + 1/9 = 4/9
4/9 + 1/27 = 12/27 + 1/27 = 13/27

it seems like sum approaches 1/2

Answer 25

yup
http://www.wolframalpha.com/input/?i=sum+from+n%3D1+to+n%3D%E2%88%9E+of+%281%2F3%29%5En

Answer 26

sounds reasonable

Answer 27

so you get it converges to?

Answer 28

yes, verified by wolfram, so my answer is (1/2) * 18 --> 9.

Answer 29

yep yep

Answer 30

tnx for helping me !

Answer 31

no problem, thanks for not just trying to get an answer

Answer 32

don't think anyone in calculus is trying to do so. Also, I don't think I will have OpenStudy on my quizes.

Answer 33

haha, well, I'm losing it over the study island high schoolers. They are driving me mad witht hte I just want an answer to my multiple guess question

Answer 34

oh, and there is one calc student just trying for an answer